if index==1: count = count_conditions( teacher_arr, lambda row: row[3] <24, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass if index==2: count = count_conditions( teacher_arr, lambda row: row[3] >24 and lambda row: row[3] <30, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass if index==3: count = count_conditions( teacher_arr, lambda row: row[3] >29 and lambda row: row[3] <35, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass if index==4: count = count_conditions( teacher_arr, lambda row: row[3] >34 and lambda row: row[3] <40, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass if index==5: count = count_conditions( teacher_arr, lambda row: row[3] >39 and lambda row: row[3] <45, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass if index==6: count = count_conditions( teacher_arr, lambda row: row[3] >44 and lambda row: row[3] <50, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass if index==7: count = count_conditions( teacher_arr, lambda row: row[3] >49 and lambda row: row[3] <55, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass if index==8: count = count_conditions( teacher_arr, lambda row: row[3] >54 and lambda row: row[3] <60, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass if index==8: count = count_conditions( teacher_arr, lambda row: row[3] >54 and lambda row: row[3] <60, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass if index==9: count = count_conditions( teacher_arr, lambda row: row[3] >=60, lambda row: row[11] == itme1, lambda row: row[8] == '专任教师') pass 优化代码
时间: 2023-06-09 12:07:32 浏览: 48
您的代码可以通过使用循环和列表,来避免重复的代码。以下是一种重构方案:
```
age_ranges = [(0, 24), (24, 30), (30, 35), (35, 40), (40, 45), (45, 50), (50, 55), (55, 60), (60, 100)]
counts = []
for i, age_range in enumerate(age_ranges):
if i == index - 1:
count = count_conditions(teacher_arr,
lambda row: age_range[0] < row[3] < age_range[1],
lambda row: row[11] == item1,
lambda row: row[8] == '专任教师')
counts.append(count)
else:
counts.append(0)
```
这样,您就可以通过循环和列表,避免重复的代码,并且可以更加清晰和易读。
相关问题
if itme1 == '总计': count = count_conditions( teacher_arr, lambda row: row[9] == '学前' or row[9] == '小学', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '女': count = count_conditions( teacher_arr, lambda row: row[2] == '女', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '少数民族': count = count_conditions( teacher_arr, lambda row: row[11] != '汉族', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '在编人员': count = count_conditions( teacher_arr, lambda row: row[12] == '是', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count代码优化
:可以考虑将重复使用的代码封装成一个函数,以减少代码重复和提高代码可读性。例如:
def count_by_criteria(teacher_arr, criteria_func, filter_func):
count = count_conditions(teacher_arr, criteria_func, filter_func)
if count != 0:
sh_4149.cell(index1, index).value = count
然后使用这个函数来代替原本重复的代码,例如:
if itme1 == '总计':
count_by_criteria(teacher_arr, lambda row: row[9] == '学前' or row[9] == '小学', lambda row: row[8] == itme)
if itme1 == '女':
count_by_criteria(teacher_arr, lambda row: row[2] == '女', lambda row: row[8] == itme)
if itme1 == '少数民族':
count_by_criteria(teacher_arr, lambda row: row[11] != '汉族', lambda row: row[8] == itme)
if itme1 == '在编人员':
count_by_criteria(teacher_arr, lambda row: row[12] == '是', lambda row: row[8] == itme)
这样代码会更易读和易维护。
for index1, itme1 in enumerate(list_col): for index, itme in enumerate(list_row): if itme1 == '总计': count = count_conditions( teacher_arr, lambda row: row[9] == '学前' or row[9] == '小学', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '女': count = count_conditions( teacher_arr, lambda row: row[2] == '女', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '少数民族': count = count_conditions( teacher_arr, lambda row: row[11] != '汉族', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '在编人员': count = count_conditions( teacher_arr, lambda row: row[12] == '是', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count代码优化
:
首先,可以将每一个if语句的判断条件都提前判断一遍,如果不满足就直接跳过,避免重复计算和赋值。
其次,可以将count_conditions函数的每个参数使用lambda函数封装成一个函数参数,以简化代码。
最后,可以将重复的赋值操作提取出来,避免重复代码。
优化后的代码如下:
count_conditions_filter1 = lambda row: row[9] == '学前' or row[9] == '小学'
count_conditions_filter2 = lambda row: row[2] == '女'
count_conditions_filter3 = lambda row: row[11] != '汉族'
count_conditions_filter4 = lambda row: row[12] == '是'
for index1, item1 in enumerate(list_col):
count = 0
if item1 == '总计':
count = count_conditions(teacher_arr, count_conditions_filter1, lambda row: row[8] == item)
elif item1 == '女':
count = count_conditions(teacher_arr, count_conditions_filter2, lambda row: row[8] == item)
elif item1 == '少数民族':
count = count_conditions(teacher_arr, count_conditions_filter3, lambda row: row[8] == item)
elif item1 == '在编人员':
count = count_conditions(teacher_arr, count_conditions_filter4, lambda row: row[8] == item)
if count != 0:
sh_4149.cell(index1 6, index 4).value = count